circumference of an ellipse


by emc92
Tags: circumference, ellipse
emc92
emc92 is offline
#1
Feb18-12, 12:05 AM
P: 33
all of my work so far is in the picture. i'm stuck on what i should do next.
Attached Thumbnails
Unknown.jpg  
Phys.Org News Partner Science news on Phys.org
Simplicity is key to co-operative robots
Chemical vapor deposition used to grow atomic layer materials on top of each other
Earliest ancestor of land herbivores discovered
genericusrnme
genericusrnme is offline
#2
Feb18-12, 05:06 AM
P: 615
your handwriting is incredibly neat, good show!
HallsofIvy
HallsofIvy is offline
#3
Feb18-12, 05:54 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879
You are making the substitution [itex]x= a sin\theta[/itex] but then your integral has both x and [itex]\theta[/itex]. That's not right.

However, I would advise using the parametric equations [itex]x= a sin(\theta)[/itex], [itex]y= b cos(\theta)[/itex] rather than that complicated equation.

emc92
emc92 is offline
#4
Feb19-12, 04:48 PM
P: 33

circumference of an ellipse


Quote Quote by HallsofIvy View Post
You are making the substitution [itex]x= a sin\theta[/itex] but then your integral has both x and [itex]\theta[/itex]. That's not right.

However, I would advise using the parametric equations [itex]x= a sin(\theta)[/itex], [itex]y= b cos(\theta)[/itex] rather than that complicated equation.
I'm not sure I understand what you mean by the parametric equations.. how does that fit into what I already have?
emc92
emc92 is offline
#5
Feb19-12, 04:48 PM
P: 33
Quote Quote by genericusrnme View Post
your handwriting is incredibly neat, good show!
thanks!
Dick
Dick is offline
#6
Feb19-12, 05:18 PM
Sci Advisor
HW Helper
Thanks
P: 25,165
Quote Quote by emc92 View Post
I'm not sure I understand what you mean by the parametric equations.. how does that fit into what I already have?
You were doing alright until after you drew the triangle. But you want to get rid of all of the x's in the thing you are integrating. And you never used dx=a cos(θ) dθ, probably because you weren't writing the dx in the integration. You need that.
emc92
emc92 is offline
#7
Feb21-12, 05:30 PM
P: 33
Quote Quote by Dick View Post
You were doing alright until after you drew the triangle. But you want to get rid of all of the x's in the thing you are integrating. And you never used dx=a cos(θ) dθ, probably because you weren't writing the dx in the integration. You need that.
oh darn! i did forget. okie well, now that I have dθ in there and i changed x^2, i still have a really ugly equation.. what should i do next?
Attached Thumbnails
Unknown.jpg  
Dick
Dick is offline
#8
Feb21-12, 05:36 PM
Sci Advisor
HW Helper
Thanks
P: 25,165
Quote Quote by emc92 View Post
oh darn! i did forget. okie well, now that I have dθ in there and i changed x^2, i still have a really ugly equation.. what should i do next?
Bring the cos(θ) inside the square root where it becomes cos^2(θ). And i) replacing the a^2 with x^2/sin^2(θ) doesn't do you any good and ii) somewhere you missed cancelling an a^2.
emc92
emc92 is offline
#9
Feb21-12, 06:07 PM
P: 33
i completely reworked it, and it looks so much better now! lol.
now i have integral from 0 to a of sqrt(1- (b^2/a^2)cos(θ))
Dick
Dick is offline
#10
Feb21-12, 06:10 PM
Sci Advisor
HW Helper
Thanks
P: 25,165
Quote Quote by emc92 View Post
i completely reworked it, and it looks so much better now! lol.
now i have integral from 0 to a of sqrt(1- (b^2/a^2)cos(θ))
I thought you were supposed to get the integral of sqrt(1-k*sin^2(θ))??
emc92
emc92 is offline
#11
Feb21-12, 06:14 PM
P: 33
right. k = 1-(b^2/a^2). did i cancel sin^2(θ) instead of cos(θ)?
Dick
Dick is offline
#12
Feb21-12, 06:23 PM
Sci Advisor
HW Helper
Thanks
P: 25,165
Quote Quote by emc92 View Post
right. k = 1-(b^2/a^2). did i cancel sin^2(θ) instead of cos(θ)?
Hard to say. What did you do?
emc92
emc92 is offline
#13
Feb21-12, 07:00 PM
P: 33
Quote Quote by Dick View Post
Hard to say. What did you do?
i've completely messed up. i don't know what to use for substitutions.. i always end up in the same place. and i don't know where dx = a cos(θ) dθ fits into all of this.

sorry i'm a lot confused!
Attached Thumbnails
Unknown.jpg  
Dick
Dick is offline
#14
Feb21-12, 09:37 PM
Sci Advisor
HW Helper
Thanks
P: 25,165
Quote Quote by emc92 View Post
i've completely messed up. i don't know what to use for substitutions.. i always end up in the same place. and i don't know where dx = a cos(θ) dθ fits into all of this.

sorry i'm a lot confused!
You are going in circles. Look you've got [itex]\int \sqrt{1+\frac{b^2 \sin^2{\theta}}{a^2 \cos^2{\theta}}} a \cos{\theta} d\theta[/itex]. Bring the cos into the square root, so you've got [itex]\int \sqrt{(1+\frac{b^2 \sin^2{\theta}}{a^2 \cos^2{\theta}}) \cos^2{\theta}} a d\theta[/itex]. Simplify inside the radical. Then use your trig identity and change the x limits to theta limits.
emc92
emc92 is offline
#15
Feb21-12, 10:24 PM
P: 33
but if k = 1 - (b^2/a^2) and under the radical says 1-ksin^2(θ), shouldn't the end result under the radical, when expanded, be 1 - (sin(θ))^2- (b^2/a^2)(sin(θ))^2?

and where does the 4 outside the integral come from?
Dick
Dick is offline
#16
Feb21-12, 10:29 PM
Sci Advisor
HW Helper
Thanks
P: 25,165
Quote Quote by emc92 View Post
but if k = 1 - (b^2/a^2) and under the radical says 1-ksin^2(θ), shouldn't the end result under the radical, when expanded, be 1 - (sin(θ))^2- (b^2/a^2)(sin(θ))^2?

and where does the 4 outside the integral come from?
No! It should be 1 - (sin(θ))^2 + (b^2/a^2)(sin(θ))^2! You are subtracting k. That's just being sloppy. And if you are integrating x from 0 to a you are only integrating over 1/4 of the ellipse. You are just covering the first quadrant.


Register to reply

Related Discussions
Spot the error! (circumference of ellipse) General Math 2
Equation of the circumference of an ellipse parametric equations Calculus & Beyond Homework 9
Circumference of an ellipse Calculus & Beyond Homework 7
Circumference of an Ellipse Introductory Physics Homework 2
circumference of an ellipse General Math 1